JEE CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

What is the minimum

\mathrm{pH}

required to prevent the precipitation of

\mathrm{ZnS}

in a solution that is

0.01 \mathrm{M} \,\, \mathrm{ZnCl}_2

and saturated with

0.10 \mathrm{M} \,\, \mathrm{H}_2 \mathrm{~S}

?
[Given

\mathrm{K}_{\mathrm{sp}}=10^{-21}, \mathrm{~K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2}=10^{-20}

]

View answer and explanation

Correct option: B

Option B (1) is correct because the minimum pH is calculated using the

K_{sp}

of ZnS, the concentration of

Zn^{2+}

, and the

K_{a1} \times K_{a2}

H_2S

More Questions from Equilibrium

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2